©2022 Ada Academica https://adac.eeEur. J. Math. Anal. 2 (2022) 6doi: 10.28924/ada/ma.2.6
Solving Equilibrium Problem and Fixed Point Problem by Normal S-iteration Process in

Hilbert Space

Shamshad Husain, Mohd Asad∗
Department of Applied Mathematics, Faculty of Engineering and Technology,

Aligarh Muslim University, Aligarh, India
s_husain68@yahoo.com, masad19932015@gmail.com
∗Correspondence: masad19932015@gmail.com

Abstract. The main purpose of this paper is to find a common element in the solution set of equilibriumproblem and fixed point problem of non-expansive mappings in the real Hilbert space with the helpof normal S-iteration process. Also, under some acceptable assumptions, we prove the sequencesinduced by above stated process converge weakly to a point in the solution set of above statedproblems. At the end, we give a numerical example to justify our work. The results studied in thiswork philosophize and boost some contemporary and known results in this direction.

1. Introduction and Auxiliary results
Everywhere in this paper except stated otherwise, let H be a real Hilbert space equipped withinner product 〈·, ·〉 and induced norm ‖ · ‖. Let C be a non-empty closed and convex subset of

H. We denote strong and weak convergence of a sequence {xn} ∈ H by the symbols → and ⇀respectively.Let T : C → H be a nonexpansive mapping. The so called fixed point problem for mapping T is tofind an element p ∈ C such that
Tp = p. (1)

Denote the set of solution of the problem (1) by Fix(T ) = {p ∈ C : Tp = p}. T is said to benonexpansive iff
‖Tp−Tq‖2 ≤‖p−q‖2, ∀p,q ∈ C.

Received: 10 Dec 2021.
Key words and phrases. equilibrium problem; normal S-iteration; fixed point problem; Hilbert space; non-expansivemapping. 1

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 2
In 2011, D.R. Sahu [4] studied problem (1) and proposed an iterative method known as NormalS-iteration Process which is defied as follows: Let x1 ∈ C be chosen arbitrarily,

yn = (1 −αn)xn + αnTxn,

xn+1 = Tyn, ∀n ≥ 1, (2)
where {αn} ⊂ (0 , 1). Under some acceptable conditions of {αn}, Sahu proved that the sequence
{xn} induced by the algorithm (2) converges weakly to an element of solution set of problem (1).The performance of normal S-iteration process is much better than Mann and Picard iterationprocess for nonexpansive mappings(see [4], [5]).Elsewhere, let F : C ×C → R be a bifunction such that for all p ∈ C, F (p,p) = 0. Then the socalled equilibrium problem is to find p ∈ C such that

F (p,q) ≥ 0, ∀q ∈ C. (3)
Denote the solution set of problem (3) by EP (F ). Problem (3) contains Nash equilibrium prob-lems, fixed point problems, variational inequality problems, minimization problems and optimizationproblems as its special cases(see [7, 16]).In this paper, we consider a problem which is formulated as follows: Find p ∈ C, such that

p ∈ Ω := Fix(T ) ∩EP (F ). (4)
In past few years, many researchers have found a common solution of problem (4) by varioustechniques(see [4], [3], [2], [12], [1], [14], [10]). Impelled and inspired by these approaches, the mainobjective of this paper is to find a common element in the solution set of problem (4) with the helpof Normal S-iteration Process in the framework of real Hilbert space. Also we prove some weakconvergence theorem under some acceptable conditions.Now we define some basic auxiliary results which are very helpful throughout this work.The metric projection PC from H into C is defined as: for any p ∈ C,

‖p−PC(p)‖≤‖p−q‖, ∀q ∈ C.

It is to be noted that the metric projection is nonexpansive. Further for any p ∈ H and s ∈ C,
s = PC(p) ⇐⇒ 〈p− s,s −q〉≥ 0, ∀q ∈ C.

A mapping T is said to be monotone iff for all p,q ∈ H
〈Tp−Tq,p−q〉≥ 0.

Lemma 1.1. [9] Let H be a Hilbert space. Then for all p,q ∈ H and α ∈ [0, 1] the followings hold:
(i) ‖p−q‖2 = ‖p‖2 −‖q‖2 − 2〈p−q,q〉;(ii) ‖p + q‖2 ≤‖p‖2 + 2〈q,p + q〉;(iii) ‖αp + (1 −α)q‖2 = α‖p‖2 + (1 −α)‖q‖2 −α(1 −α)‖p−q‖2.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 3
Assumption 1.1. [6] Let F : C ×C →R be a bi-function satisfying the subsequent conditions:(i) F (p,p) ≥ 0, ∀p ∈ C;(ii) F is monotone, i.e. F (p,q) + F (q,p) ≤ 0, ∀p,q ∈ C;(iii) F is upper semi continuous, i.e. for each p,q,s ∈ C,

lim
t→0

supF (λs + (1 −λ)p,q) ≤ F (p,q); (5)
(iv) For each fixed p ∈ C, the function q 7→ F (p,q) is convex and lower semi continuous;
Lemma 1.2. [7] Assume that the bi-function F : C ×C → R satisfy the conditions of Assumption1.1. Then for fixed r > 0 and p ∈ H, there exists s ∈ C such that

F (q,p) +
1

r
〈q −p,p− s〉≥ 0, ∀q ∈ C. (6)

Lemma 1.3. [12] Assume that the bi-function F : C ×C →R satisfy the conditions of Assumption1.1. If for r > 0 and p ∈ H, defined a mapping TFr : H → C as follows:
TFr (p) =

{
s ∈ C : F (s,q) +

1

r
〈q − s,s −p〉≥ 0, ∀q ∈ C

}
. (7)

Then the followings hold:
(i) TFr is non-empty and single valued.(ii) TFr is firmly non-expansive, i.e.,

‖TFr (p) −T
F
r (q)‖

2 ≤〈TFr (p) −T
F
r (q),p−q〉 ∀p,q ∈ H.

(iii) Fix(TFr ) = EP(F ).(iv) EP(F ) is closed and convex.
Lemma 1.4. [11] Let {an} be a sequence of non negetive real numbers such that

an+1 ≤ (1 −αn)an + αnδn + γn, ∀n ≥ 0,

where αn ∈ (0, 1) and δn ⊂R satisfies the following conditions:(i) ∑∞n=0αn = ∞;(ii) lim
n→∞

supδn ≤ 0.(iii) γn ≥ 0 (n ≥ 1),∑γn < ∞.Then lim
n→∞

an = 0.

Lemma 1.5. [13] Let C be a closed and convex subset of H and T : C → C be a non-expansivemapping. Then(i) Fix(T ) is a closed and convex subset of C;(ii) I −T is demiclosed at 0.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 4
Lemma 1.6. [8] Let F : C ×C →R be a non linear bi-function satisfying the Assumption 1.1 andlet TFr be defined as above in Lemma 1.3. If for r > 0, let p,q ∈ H and r1, r2 > 0, Then

‖TFr2 (q) −T
F
r1

(p)‖≤‖q −p‖ +
∣∣∣∣r2 − r1r2

∣∣∣∣‖TFr2 (q) −q‖.
Lemma 1.7. [15] Let xn and yn be two bounded sequences in a Banach space X and let βn be asequence in [0, 1] which satisfy the following conditions:

0 < lim
n→∞

inf βn ≤ lim
n→∞

sup βn < 1.

Suppose xn+1 = (1 −βn)zn + βnxn for all integers n ≥ 0, and
lim
n→∞

sup (‖zn+1 − zn‖−‖xn+1 − xn‖) ≤ 0, Then lim
n→∞

‖xn − zn‖ = 0.

2. Main Result
In this section we study and analyze Normal S-iteration process for solving equilibrium problemand fixed point problem for nonexpansive mapping and its convergence analysis.

Theorem 2.1. Let C ⊂ H be a nonempty closed and convex subsets of H. Let F : C ×C → R bea nonlinear bifunction satisfying Assumption 1.1. Let T : C → H be a nonexpansive mapping suchthat Fix(T ) 6= . Assume that Ω := Fix(T ) ∩EP (F ) 6= . Let {xn}be a sequence defined as follows:Choose x1 ∈ H arbitrarily,
yn = T

F
rn

(xn),

zn = (1 −αn)yn + αnTyn,

xn+1 = Tzn, ∀n ≥ 1, (8)
where {αn}⊂ [0 , 1] and {rn}⊂ (0 ,∞) satisfying the following conditions:

C1: lim
n→∞

αn = 0,
∑∞
n=1αn(1 −αn) = ∞,

∑∞
n=1 |αn −αn−1| < ∞;

C2: lim
n→∞

inf rn > 0,
∑∞
n=0 |rn+1 − rn| < ∞;Then the sequence {xn} induced by process (8) converges weakly to an element in Ω.

Proof. Take p ∈ Ω. Then by process (8), we obtain
‖xn+1 −p‖ = ‖Tzn −p‖≤‖zn −p‖,

≤‖(1 −αn)yn + αnTyn −p‖,

≤ (1 −αn)‖yn −p‖ + αn‖Tyn −p‖,

≤ (1 −αn)‖yn −p‖ + αn‖yn −p‖,

≤‖yn −p‖≤‖TFrn (xn) −p‖,

≤‖xn −p‖.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 5
By using mathematical induction, we have

‖xn+1 −p‖≤‖xn −p‖≤‖x1 −p‖, ∀n ≥ 1.

Hence the sequence {xn} is bounded and so are the sequences {yn},{zn},{Tyn} and {Tzn} arealso bounded.Let M = supn≥0{‖yn −xn‖ + ‖xn −q‖2 + ‖Tyn‖ + ‖Tzn‖}.Since yn = TFrn (xn) and yn−1 = TFrn−1(xn−1), Then we obtain
F (yn,q) +

1

rn
〈q −yn,yn −xn〉≥ 0, ∀q ∈ C, (9)

F (yn−1,q) +
1

rn−1
〈q −yn−1,yn−1 −xn−1〉≥ 0, ∀q ∈ C. (10)

Replace q by yn in (10) and q by qn−1 in (9) and adding them with the Assumption 1.1(ii),we obtain
〈yn −yn−1,

yn−1 −xn−1
rn−1

−
yn −xn
rn

〉≥ 0,

and hence
〈yn −yn−1,yn−1 −yn −xn−1 −

rn−1
rn

(yn −xn)〉≥ 0.

This implies that by using Lemma 1.6
‖yn −yn−1‖2 ≤〈yn −yn−1,xn −xn−1 +

(
1 −

rn−1
rn

)
(yn −xn)〉,

≤‖yn −yn−1‖
{
‖xn −xn−1‖ +

∣∣∣∣rn − rn−1rn
∣∣∣∣‖yn −xn‖},

‖yn −yn−1‖≤‖xn −xn−1‖ +
∣∣∣∣rn − rn−1rn

∣∣∣∣‖yn −xn‖,
From process (8)(C2), we have lim

n→∞
inf rn > 0. Therefore there exists r > 0 such that rn > r forlarge enough n ∈N. Then for n ≥ 1,

‖yn −yn−1‖≤‖xn −xn−1‖ +
1

r
|rn − rn−1|M. (11)

Consider
‖xn+1 −xn‖ = ‖Tzn −Tzn−1‖≤‖zn −zn−1‖,

≤‖(1 −αn)yn + αnTyn − (1 −αn−1)yn−1 −αn−1Tyn−1‖,

≤‖(1 −αn)yn − (1 −αn)yn−1 + (1 −αn)yn−1 − (1 −αn−1)yn−1

+ αnTyn −αnTyn−1 + −αnTyn−1 −αn−1Tyn−1‖,

≤ (1 −αn)‖yn −yn−1‖ + 2|αn −αn−1|M + αn‖yn −yn−1‖,

≤‖yn −yn−1‖ + 2|αn −αn−1|M. (12)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 6
Using (11) and (12), we obtain

‖xn+1 −xn‖≤‖xn −xn−1‖ +
1

r
|rn − rn−1|M + 2|αn −αn−1|M. (13)

By applying Lemma 1.4, we obtain
lim
n→∞

‖xn+1 −xn‖ = 0. (14)
By using process (8)(C1)(C2) along with Lemma 1.7 and (13), we obtain

lim
n→∞

‖xn −zn‖ = 0. (15)
Furthermore, for any p ∈ Ω, we have from process (8)

‖yn −p‖2 = ‖TFrn (xn) −p‖
2,

≤〈TFrn (xn) −T
F
rn

(p),xn −p〉,

≤〈yn −p,xn −p〉,

≤
1

2

{
‖yn −p‖2 + ‖xn −p‖2 −‖xn −yn‖2

}
,

≤‖xn −p‖2 −‖xn −yn‖2. (16)
From convaxity of function x 7→ ‖x‖2 and (16), we obtain

‖xn+1 −p‖2 = ‖Tzn −p‖2,

≤‖zn −p‖2,

≤‖(1 −αn)yn + αnTyn −p‖2,

≤ (1 −αn)‖yn −p‖2 + αn‖Tyn −p‖2,

≤‖yn −p‖2,

≤‖xn −p‖2 −‖xn −yn‖2.

And so,
‖xn −yn‖2 ≤‖xn −p‖2 −‖xn+1 −p‖2,

≤ (‖xn −p‖−‖xn+1 −p‖)(‖xn −p‖ + ‖xn+1 −p‖),

≤‖xn −xn+1‖(‖xn −p‖ + ‖xn+1 −p‖).

Since the sequence {xn} is bounded and lim
n→∞

‖xn+1 −xn‖ = 0. We have
lim
n→∞

‖xn −yn‖ = 0. (17)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 7
Further,

‖xn+1 −p‖2 = ‖Tzn −p‖2,

≤‖zn −p‖2,

≤‖(1 −αn)yn + αnTyn −p‖2,

≤ (1 −αn)‖yn −p‖2 + αn‖Tyn −p‖2 −αn(1 −αn)‖yn −Tyn‖2,

≤‖yn −p‖2 −αn(1 −αn)‖yn −Tyn‖2,

≤‖xn −p‖2 −‖xn −yn‖2 −αn(1 −αn)‖yn −Tyn‖2,

and so,
αn(1 −αn)‖yn −Tyn‖2 ≤‖xn −p‖2 −‖xn+1 −p‖2 −‖xn −yn‖2,

≤‖xn −xn+1‖(‖xn −p‖ + ‖xn+1 −p‖) −‖xn −yn‖2,

using process (8)(C1), (14) and (17), we obtain
lim
n→∞

‖yn −Tyn‖ = 0. (18)
consider

‖zn −Tzn‖≤‖zn −yn‖ + ‖yn −Tyn‖ + ‖Tyn −Tzn‖, (19)
≤‖zn −yn‖ + ‖yn −Tyn‖ + ‖yn −zn‖. (20)

By using (15) and (18), we obtain
lim
n→∞

‖zn −Tzn‖ = 0. (21)
Since {xn} is bounded. There exists a subsequence {xni} ⊂ {xn} such that xn ⇀ p̂. Since
lim
n→∞

‖xn −yn‖ = 0 and {yn} is bounded, this implies that yni ⇀ p̂ ∈ C. Now by (18) we have
‖Tyni −yni‖→ 0. (22)

From (22) and Lemma 1.5, we conclude that p̂ ∈ Fix(T ).Next we prove that p̂ ∈ EP (F ). Since yn = TFrn (xn), we have
F (yn,q) +

1

rn
〈q −yn,yn −xn〉≥ 0, ∀q ∈ C.By using Assumption 1.1(ii), we obtain

1

rn
〈q −yn,yn −xn〉≥ F (q,yn),

and so,
〈q −yni,

yni −xni
rni

〉≥ F (q,yni ). (23)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 8
Since ‖yni−xni‖

rni
≤ ‖yni−xni‖

r
→ 0 and yni ⇀ p̂, therefore by Assumption 1.1(iv), we obtain

lim
ni→∞

inf F (q,yni ) ≤ lim
ni→∞

〈q −yni,
yni −xni
rni

〉 = 0.

That is,
F (q, p̂) ≤ 0, ∀q ∈ C. (24)

Further for any λ ∈ (0 , 1) and q ∈ C, let qλ = λq + (1 − λ)p̂, then qλ ∈ C and so we have
F (qλ, p̂) ≤ 0. It follows from the Assumption 1.1 and (24), that

0 = F (qλ,qλ),

≤ λF (qλ,q) + (1 −λ)F (qλ, p̂),

≤ λF (qλ,q).

This implies that F (qλ,q) ≥ 0, ∀λ ∈ (0 , 1). Letting λ → 0+ by Assumption 1.1, we have
F (p̂,q) ≥ 0, ∀q ∈ C. This implies that p̂ ∈ EP (F ) and hence p̂ ∈ Ω. This completes theproof. �

3. Numerical Example
Here we give numerical examples for supporting our main results. All codes are done by Matlab2021a.

Example 3.1. Set H = R. Let C = [0 + ∞). Suppose T : C → H, is defined by T (p) = p
3
. It

can be easily seen that, here Fix(T ) = {0}. Also, we define F (s,q) = 3q2 + 2sq − 5s2, It is easy
to check that F satisfy the conditions of Assumption 1.1. So, for rn = r > 0, TFr (p) is non-empty
and single-valued for each p ∈ C. Hence for r > 0, there exists s ∈ C such that

F (s,q) +
1

r
〈q − s,s −p〉≥ 0 ∀ q ∈ C,

which is equivalent to

3rq2 + (s −p + 2rs)q + (ps − 5rs2 − s2) ≥ 0, ∀ q ∈ C.

After solving the above inequality, we get s = p
1+8r

for each r > 0 i.e. TFr (p) =
p
1+8r

for each
r > 0. It can be easily seen that here EP (F ) = {0}. This implies that Ω := Fix(T )∩EP (F ) = {0}.
Now, let us choose r = 1

8
, and {αn} = 1(n+6). {αn} satisfy the conditions of main result.

Table. For different initial value, we present a table of iterations here.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.6 9
No. of iterations x0 = 1 x0 = −1

1 1.000000 -1.0000002 0.150794 -0.1419233 0.023038 -0.0204074 0.003555 -0.0029645 0.000553 -0.0004346 0.000087 -0.0000647 0.000014 -0.0000098 0.000002 -0.0000019 0.000000 0.000000

0 2 4 6 8 10 12 14 16 18 20

Number of iterations

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

X
n

x
0
 = 1

x
0
 = -1

Figure 1. Graphical representation of sequence {xn} for different choices of initialvalue x0.
References

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	1. Introduction and Auxiliary results
	2. Main Result
	3. Numerical Example
	References